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A Novel Look at the One Dimensional 
Delta Schrodinger Equation 

By 

Patrick Bruskiewich 

Department of Physics and Astronomy, 

University of British Columbia 

email: patrickb@physics . ubc . ca 

Resume 

In this paper both integral transforms and an algebraic reformulation 
of the second order differential equation are used to derive the solution to 
the one dimensional attractive delta Schrodinger equation. This is a novel 
approach not found in any standard textbook. Both the transform and 
the reformulation methods provide additional insight into the nature of 
the one dimensional attractive delta potential. 

Introduction 

In many problems in quantum mechanics one can arrive at a solution in 
several ways. It is sometimes interesting to revisit a familar problem to 
see what new can be learned. Both algebraic and analytic techniques are 
used in this paper to solve for the one dimensional Schrodinger equation 
with an attractive delta potential. 

An attractive delta potential can be used to model a point particle, one 
dimensional models of molecules with multiple centres and in the Kronig- 
Penney model of crystalline structure to name just a few occurences of the 
potential in the literature. It is also of central importance in the Green's 
Function formulation of quantum mechanics. 



22 



Gamma 138 Patrick Bruskiewich 

The traditional manner in which to work with the one dimensional 
Schrodinger equation is through differentiation. The fundamental equa- 
tions of quantum mechanics can also be derived from a variational prin- 
ciple or by means of integration. This has direct bearing on the manner 
in which the solution to the one dimensional attractive delta Schrodinger 
equation is derived in this paper. 

We must begin by being mindful that the delta function is not a func- 
tion in the traditional sense. It is a generalized function, or distribution, 
which can be thought of as a limit of a sequence of functions of ever 
increasing height and ever-decreasing width, used as an integrand in an 
integral equation. 

For example in its simplest representation, we can model an attractive 
delta potential by a square well potential in the limit that its depth (Vo) 
approaches infinity while its width (a) approaches but never reaches zero 
in the limit, (with the proviso that Vo a ~ 1 in the limit). 

In the limit we can also represent a delta function in several other 
relevant ways, namely 



s{x) = 7?^7? exp[ -7 

5(x) = — lim 



7T x->0 x 2 + e 2 

_, . 1 siniVa: 
dlx) = — lim 

As is evident, each of these representations has a characteristic width 
which approaches, but never reaches zero in their limit. This is of impor- 
tance when we remember that a physical state must always correspond 
to a square-integrable wavefunction. 

If 6(x — a) represents a delta function at point a then given some 
function f(x) 

23 



A Novel Look at the One Dimensional Delta Schrodinger Equation Gamma 138 



/OO rOO 

f(x)5(x — a)dx = fia) / 6(x — a)dx = fid) 
-00 ' J — OO 

The standard interpretation of the delta function, irrespective of its 
functional representation, is that under the integral sign the delta func- 
tion serves to "pick out" the value of the function fix) at the point a. 

The Traditional Derivation 

Consider the one dimensional Schrodinger equation with an attractive 
delta potential given by 



H 2 d 2 

ip(x) — a 6(x) ip(x) = E ift(x). 



2m dx 2 



The standard way to solve this equation is to match continuity condi- 
tions. For a potential centred on x = that spans — oo < x < +oo we 
consider both the Dirichlet boundary condition and the Neumann bound- 
ary condition. [1] 

Let us consider the representation of the delta function as a box of 
small width 2e centred on x = and large negative potential —a. In the 
region x < — e, the potential is zero so we have a solution of the form 



ift-(x) = Aex.p(nx) 



where k = ^ — ^ . 

Similarly in the region x > e the potential is again zero and in this 
region we have a solution of the form 



ip+(x) = .Bexp(— kx) 
24 



Gamma 138 Patrick Bruskiewich 



It remains only to match these two functions at x = with the bound- 

dx 



ary conditions, namely ip is continuous and -f- is continuous except at 



points where the potential is infinite. 

Evidently the delta function must describe the discontinuity in the 
derivative of ip at x = 0. We can extract the delta function by integrating 
the Schrodinger equation, from — e to +e, and then take the limit as e — > 0, 
namely 

/ !T~2 ^( x ) dx — a 5(x) ip(x) dx = E ^{x) dx. 

In the limit the last integral is clearly equal to zero, while the first 
integral is 

/: S «•> - - *2> 

which is a measure as to how much the slope in the wavefunction changes 
over the interval — e < x < e. 

Thus we have simplified the expression to 



.,dib. —2ma .. r+e —2ma , , n . 

A( i ) = -&- H L *« *w dx = n^ W) 

Through integration we have reduced the order of the Schrodinger 
equation from second to first order and we incorporated the boundary 
conditions in the first order equation explicitly. 

By symmetry we recognize that 



Solving this first order differential equation we arrive at the familiar 
normalized wavefunction solution given by 

25 



A Novel Look at the One Dimensional Delta Schrodinger Equation Gamma 138 



, . y/2ma . ma I x I 

* (x) = -jr~ exp( — w~ ] 

The energy of this normalized wavefunction solution is given by 

ma 2 

£ = ~w 

There are several other ways in which to derive the solution to the one 
dimensional Schrodinger equation with an attractive delta potential. The 
alternate techniques outlined in this paper can act as a starting point to 
describing the quantum behaviour of other systems. 

Solution by an Integral Transform 

Consider once again the one dimensional attractive delta Schrodinger 
equation 



h 2 d 2 

ip — a 5(x) tjj = E tp. 



2m dx 2 



We can borrow a technique from quantum field theory and derive a so- 
lution using an integral transform in the way of a Fourier transform. [2] 
Let 



1 °r 

^(x) = — dk e lkx cf>(k) 

— 00 



where <j){k) is the wave function in the momentum representation. Intro- 
ducing this solution into the differential equation we find that 



(^ k 2 + B) m = a V>(0) 
26 



Gamma 138 



Patrick Bruskiewich 



where "0(0) is the wavefunction value at x = and B 
of the eigensolution. Solving for <j){k) we find 



E, is the energy 



m 



a 



*(0) 



B 2 + B 



Using the completeness theorem, 



00 00 00 

1 = I $*{x) if)(x) dx = J J 



dk dk' 



J(k-k/)x ±*( h i 



— 00 



— 00 —00 



2tt 2tt 



WW*) 



which when integrated over k' simplifies to 



-i 00 

— / dk (f)*(k) <f>(k) 

— 00 



2maip(ti) 

h 2 



-i 00 

- /- 

7T i 



dk 



[fc 2 + 



2mB ] 2 



where we recognize that the integrand is an even function and change 
the integral limits accordingly. Upon integration we arrive at an inverse 
tangent function, namely 



tf 



2maijj(0)_ 



2 7T 



n i 3 r kh - k=0 ° 

arctan 



V2mB 



V2mB 



k=0 



which when we solve for E yields (E = — B) 



E = - 



ma 



2h 2 



We can use our value for B in the wavefunction (j){k) and solve for ip(x) 
in the transform equation, namely 



i/)(x) = 



2ma V>(0) 

h 2 



°r dk 



-Akx 



-oo 



2tt k 2 +[ 



ti 2 



27 



A Novel Look at the One Dimensional Delta Schrodinger Equation Gamma 138 

By inspection we see that only the even function cos(A;a:) yields a non- 
zero contribution, and so adjusting the limits of integration accordingly 
we find 



*(*) = — si— J- 



2ma "0(0) °f dk cos(kx) 

k 2 + [f^ 12 

which yields the solution 



il){x) = V>(0)exp( — — ) 

Solution by Reformulation 

We can find a solution to the one dimensional attractive delta Schrodinger 
equation by using a simple transformation to the second order differential 
equation. [3] 

Consider the delta function S(x) represented by the second partial 
derivative of the absolute value function in x, namely 



1 d 2 
*(*) = 2W I X I • 

Gathering the terms in the second derivative we can rewrite the Schrodinger 
equation as 

h " d " ! a r Q2 i 11/ w , 



2m dx 2 2 dx 2 

Bringing the wavefunction ip over to the left hand side, we get 



1 d ,dib s ma d ,d\x\ s 2m I E 



ip dx dx h 2 dx dx h 2 

28 



Gamma 138 Patrick Bruskiewich 

We have used the fact that for a bounded wavefunction the energy E 
is less than zero, namely 

E = -\E\. 

Bringing the wavefunction ip into the bracketed derivative and gather- 
ing terms we have 

d , 1 dip ma d I x L 1 , dip . 9 2ra I E I 

— ( — H ! — -) H (— -) = ! -■ 

dx ip dx H 2 dx ip 2 dx h 2 

Let us set the following condition on tp 

d A dip ma d \ x \ 
dx ip dx h 2 dx 
Solving this equation we get 

^ = ■0(0) exp( — — | x |). 

Using the normalization condition we find the familiar wavefunction 
for the delta potential 

. J ma , ma . .. 

^ = __ exp( __ | x p. 

Having found ip we can now use this wavefunction to solve for the 
remaining condition, namely the value for the energy of the wavefunction, 



1 M.n 2m I E 
"(-ST) = 



ip 2 dx h 2 

Solving for the Energy (| E |= — E we get 



„ ma 2 

E = 



2h 2 ' 

which is the expected value for the energy of a bounded wavefunction for 
the delta potential. 



29 



A Novel Look at the One Dimensional Delta Schrodinger Equation 



Gamma 138 



Solution by a Laplace Transform 

Consider once again the one dimensional attractive delta Schrodinger 
equation 



2 £>2 



K l d 



2m dx 2 



ip — a 5(x) ip = E i\). 



We can solve for this second order differential equation using the Laplace 
transform. Applying the usual transform 



00 



L(f) = Jf(t)e- st dt 



to this second order differential equation yields 

( s 2 + -ft- ) L (VO + ( -^- - s) V^(O) - <//(0) = 
Let i/j(0) = C and ^'(0) = 7 so that 



9 2mE 2ma 

" ' ) L (VO + ( -r^- - s ) C - 7 



(* 2 + 



ft 2 



ft 2 







Solving for L (*0) provides the following transform equation which 
needs to be solved: 



MVO 



(7-^0 

( s 2 + b 2 ) ' ( s 2 + b 2 ) 



*c 



+ 



where 6 2 = -^-. The inverse transform of the first term is 



L- 1 



(s 2 + b 2 ) 



( cos ibx) 



30 



Gamma 138 



Patrick Bruskiewich 



If we let r\ = (7 — ^p C) then the inverse transform of the second 
term is of the form 



r\ L 



-1 



2ib (s-ib) 2ib (s + ib) 



V 



2ib 



kl 



Jbx 



—ibx 



] 



1 

b 



sin 



(bx) 



So the solution for a one dimensional attractive delta Schrodinger equa- 
tion using the Laplace transform method is 



tjj(x) 



Tj 

C cos (bx) + — sin (bx) 



( cos (bx) + 



1 



7 



2ma 



c 



sin (bx) 



which appears of a very different form to that of the solution given in part 
one of this paper. However, it is possible to rewrite this eigensolution in 
a more familiar form. 



If we have a bounded solution we can change E — > 



E I giving 



ip(x) = £ cosh (| b I x) — 
= C cosh (| b I x) + 



77 



sinh (| b I x) 



7 



2ma 



c 



sinh (| b I x) 



since b -> i \ b |, cos (bx) — > cosh (| b \ x) and sin (bx) — >• i sinh (| 6 | #) 
under this tranformation. 

Since we are in the domain < x < 00 in our solution we need to 
reformulate that part of cosh (bx) to conform to our boundary conditions 
as # — > 00. We can do this by recognizing that 



C exp (— I b I x) = ( [ cosh (| b \ x) — sinh (| b \ x) 



m \a\ 



At this point, analysis of the sinh (bx) term shows that \ b \ n , 
that this term can be rewritten as 



so 



31 



A Novel Look at the One Dimensional Delta Schrodinger Equation 



Gamma 138 



1 

1 b\ 


7 - 


2ma 

h 2 C J 


7 - 


ma 

h 2 C J 


sinh (| 



sinh (| b | x) = 



sinh (| b | x) + (, sinh (| b \ x) 



We see then we have 



tp(x) = £ exp (— | b | x) + 



ma 



c 



7 



sinh (| b | x) 



where the first term is the familiar solution to the one dimensional at- 
tractive delta Schrodinger equation. There is still the second term, the 
sinh (| b | x) part of the solution, we need to interpret. 

Recalling that ift'(0) = 7 we see that the term in sinh(| b \ x) is, in 
fact 



1 

T 



ma 



C - 7 



sinh (| b I x) = 



c - 



<//(0) 



sinh (| b I x) 



By inspection we know that as x — > 00, the function sinh(| b \ x) — > 
00. With some thought, however, and a careful consideration of the sign, 
in light of the | b \ term in the expression 



c 



V/(0) 



one sees immediately that we recover the well known solution to the 
one dimensional Schrodinger equation with an attractive delta potential, 
namely 



, . y/2ma , ma I x L 
tp{x) = — r — exp( — ) 



h 



h 2 



32 



Gamma 138 Patrick Bruskiewich 

Discussion 

The algebraic reformulation outlined above is a novel approach not found 
in the literature, as is the Laplace Transform approach. The use of the 
Laplace transform to solve one dimensional equations in quantum me- 
chanics is an under-utilized technique which should be explored fur- 
ther. The direct mathematical tie between the Laplace transform and 
the Fourier transform method is not so obvious, but is rich in implication. 

Along with having applicability in quantum mechanics, the one dimen- 
sional attractive delta potential is a good starting point to the study of 
the Dirac delta potential in more than one dimension. This multidimen- 
sional potential in turn allows the introduction within the framework of 
quantum mechanics of many of the basic concepts of quantum field the- 
ory, such as regularization, the renormalization group and dimensional 
transformation. 

References 

[1] The standard derivation of the solution to the one dimensional attrac- 
tive delta Schrodinger equation can be found in most any quantum 
textbook, such as the texts by Dirac, Messiah, Merzbacher, Grif- 
fiths or Gasiorowicz. 

[2] For a succinct overview of this technique refer to section 10.8 of the 
book Quarks, Leptons and Gauge Fields by Kerson Huang, World 
Scientific, Singapore, 1982. 

[3] The idea for this approach derives from a comment in section 6.2 
of Quantum Mechanics, 2nd Edition by Eugen Merzbacher, John 
Wiley and Sons, New York, 1970. 



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